| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area enclosed by polar curve |
| Difficulty | Standard +0.8 This is a multi-part polar coordinates question requiring sketching, finding tangents at the pole (requiring knowledge that these occur when r=0), identifying symmetry, converting between polar and Cartesian coordinates, and applying the polar area formula. While the techniques are standard for FP2, the combination of multiple concepts and the need to work with r=sin(5θ) (a rose curve) makes this moderately challenging, above average difficulty but not requiring exceptional insight. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Single enclosed loop in 1st quadrant, tangent at origin | B1 | Single enclosed loop in 1st quadrant, tangent at origin (not necessarily seen) |
| Less than \(\theta = \frac{\pi}{4}\) | B1 | |
| \(\theta = \frac{\pi}{5}\) stated | B1 | |
| \(\theta = 0\) stated; −1 any extra "tangents" if 2 are correct | B1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\theta = \frac{\pi}{10}\) | B1 | For \(\theta\). Allow Cartesian \(y = \left(\tan\frac{\pi}{10}\right)x\) |
| \(x = r\cos\theta = \cos\frac{\pi}{10}\ (\approx 0.951)\), \(y = r\sin\theta = \sin\frac{\pi}{10}\ (\approx 0.309)\) | B1 | For both \(x\) and \(y\). Allow decimal values to 3sf or better. [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A = \frac{1}{2}\int_0^{\pi/5} r^2\,d\theta = \frac{1}{2}\int_0^{\pi/5}\sin^2 5\theta\,d\theta\) | M1 | Correct formula for area with correct limits; or \(A = \int_0^{\pi/10} r^2\,d\theta\) |
| \(= \frac{1}{4}\int_0^{\pi/5}(1-\cos 10\theta)\,d\theta\) | M1 | Correct method to get integrand |
| \(= \frac{1}{4}\left[\theta - \frac{1}{10}\sin 10\theta\right]_0^{\pi/5}\) | A1 | Dep on 2nd M. Integral — ignore limits |
| \(= \frac{1}{4}\left(\frac{\pi}{5} - \frac{1}{10}\sin 2\pi\right) = \frac{\pi}{20}\) | A1 | [4] |
## Question 6:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Single enclosed loop in 1st quadrant, tangent at origin | B1 | Single enclosed loop in 1st quadrant, tangent at origin (not necessarily seen) |
| Less than $\theta = \frac{\pi}{4}$ | B1 | |
| $\theta = \frac{\pi}{5}$ stated | B1 | |
| $\theta = 0$ stated; −1 any extra "tangents" if 2 are correct | B1 | [4] |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\theta = \frac{\pi}{10}$ | B1 | For $\theta$. Allow Cartesian $y = \left(\tan\frac{\pi}{10}\right)x$ |
| $x = r\cos\theta = \cos\frac{\pi}{10}\ (\approx 0.951)$, $y = r\sin\theta = \sin\frac{\pi}{10}\ (\approx 0.309)$ | B1 | For both $x$ and $y$. Allow decimal values to 3sf or better. [2] |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $A = \frac{1}{2}\int_0^{\pi/5} r^2\,d\theta = \frac{1}{2}\int_0^{\pi/5}\sin^2 5\theta\,d\theta$ | M1 | Correct formula for area with correct limits; or $A = \int_0^{\pi/10} r^2\,d\theta$ |
| $= \frac{1}{4}\int_0^{\pi/5}(1-\cos 10\theta)\,d\theta$ | M1 | Correct method to get integrand |
| $= \frac{1}{4}\left[\theta - \frac{1}{10}\sin 10\theta\right]_0^{\pi/5}$ | A1 | Dep on 2nd M. Integral — ignore limits |
| $= \frac{1}{4}\left(\frac{\pi}{5} - \frac{1}{10}\sin 2\pi\right) = \frac{\pi}{20}$ | A1 | [4] |
---6 The equation of a curve in polar coordinates is $r = \sin 5 \theta$ for $0 \leqslant \theta \leqslant \frac { 1 } { 5 } \pi$.\\
(i) Sketch the curve and write down the equations of the tangents at the pole.\\
(ii) The line of symmetry meets the curve at the pole and at one other point $A$. Find the equation of the line of symmetry and the cartesian coordinates of $A$.\\
(iii) Find the area of the region enclosed by this curve.
\hfill \mbox{\textit{OCR FP2 2016 Q6 [10]}}